Notes · Further MathematicsUK · A-Levels

Polar coordinates

Polar coordinates locate a point by its distance $r$ from a fixed pole and the angle $\theta$ its radius makes with a fixed initial line. This topic covers converting between polar and Cartesian forms, sketching polar curves such as circles, cardioids and roses, finding tangents parallel and perpendicular to the initial line, and computing the area enclosed by a polar curve with the formula $\tfrac{1}{2}\int r^2\,\mathrm{d}\theta$ .

4 sections~11 min reading time3 competenciesLevel Standard 1 · Advanced 3

T·0777 / 13

Exam profile

AO1 · Convert between polar and Cartesian forms and evaluate polar areas accuratelyAO2 · Derive the polar area formula and sketch curves from their polar equationsAO3 · Set up and evaluate areas for composite polar regions and loops

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basic level

AS Further Mathematics introduces polar coordinates, conversion and simple curve sketching.

higher level

The full A-Level adds tangents to polar curves and the calculation of areas, including areas of loops and between curves.

Depth

Text

Contents · 4 sections

Polar coordinates

Polar coordinates and conversion#

StandardAQA 7367 §3.7DfE Further Mathematics

The polar curve r = 2 cos θ

Fig. 1The equation

r = 2\cos\theta

traces a circle of radius

1

centred at

(1,0)

, passing through the pole.

In the polar system a point is given by

(r, \theta)

, where

r

is its distance from the pole (the origin) and

\theta

is the angle, measured anticlockwise, that the line from the pole to the point makes with the initial line (the positive

x

-axis). At this level the convention is that

r \geq 0

and

\theta

is taken in a suitable range such as

0 \leq \theta < 2\pi

-\pi < \theta \leq \pi

Conversion between polar and Cartesian coordinates uses the right-angled triangle formed by the point and its projections:

x = r\cos\theta

and

y = r\sin\theta

go from polar to Cartesian, while

r = \sqrt{x^2 + y^2}

and

\tan\theta = \dfrac{y}{x}

go the other way. As with the argument of a complex number, finding

\theta

requires attention to the quadrant, using the signs of

x

and

y

, not just the value of

\arctan(y/x)

Equations of curves convert too, and this is often the quickest way to identify a polar curve. Multiplying a polar equation by

r

to create

r^2

r\cos\theta

r\sin\theta

lets you substitute

r^2 = x^2 + y^2

x = r\cos\theta

and

y = r\sin\theta

. For instance

r = 2\cos\theta

becomes

r^2 = 2r\cos\theta

, i.e.

x^2 + y^2 = 2x

, which rearranges to

(x-1)^2 + y^2 = 1

— a circle of radius

1

centred at

(1,0)

passing through the pole.

This interplay is powerful: a curve that looks unfamiliar in one system may be elementary in the other. A straight line, a circle through the pole and a circle centred at the pole all have simple polar equations, and recognising these standard forms —

r = a

(circle centred at the pole),

r = 2a\cos\theta

(circle through the pole) and

\theta = \alpha

(a half-line from the pole) — lets you sketch and interpret polar curves quickly.

x = r\cos\theta, \quad y = r\sin\theta, \qquad r = \sqrt{x^2 + y^2}, \quad \tan\theta = \frac{y}{x}

Polar-Cartesian conversion

Determine $\theta$ from the quadrant using the signs of $x$ and $y$ .

r = 2a\cos\theta \ \Longleftrightarrow\ (x-a)^2 + y^2 = a^2

Circle through the pole

A circle of radius $a$ centred at $(a,0)$ , passing through the origin.

Worked example

Converting a polar equation

Convert $r = 4\sin\theta$ to Cartesian form and describe the curve.

Multiply by r
$r^2 = 4r\sin\theta$ .
Substitute
Using $r^2 = x^2 + y^2$ and $r\sin\theta = y$ : $x^2 + y^2 = 4y$ .
$x^2 + y^2 = 4y$
Complete the square
$x^2 + (y^2 - 4y + 4) = 4$ , i.e. $x^2 + (y-2)^2 = 4$ .
Describe
A circle of radius $2$ centred at $(0, 2)$ , passing through the pole.

Result: $x^2 + (y-2)^2 = 4$ : a circle, radius $2$ , centre $(0,2)$ .

Typical mistakes

Getting $\theta$ out by $\pi$ by ignoring the quadrant when converting from Cartesian to polar.
Misidentifying $r = 2a\cos\theta$ as a curve centred at the pole; it is a circle through the pole centred at $(a,0)$ .

Active revision

Convert $r = 4\sin\theta$ to Cartesian form and describe the curve, and convert the point $(x,y) = (-1, \sqrt{3})$ to polar form with $0 \leq \theta < 2\pi$ .

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Sketching polar curves#

AdvancedAQA 7367 §3.7DfE Further Mathematics

The cardioid r = 1 + cos θ

Fig. 2The cardioid reaches its maximum radius

2

along the initial line and has a cusp at the pole, symmetric about the

x

-axis.

To sketch

r = f(\theta)

, build a picture of how the radius grows and shrinks as

\theta

sweeps round. A short table of

r

at key angles (

0, \tfrac{\pi}{4}, \tfrac{\pi}{2}, \dots

) reveals where the curve is farthest from and nearest to the pole, and where it touches the pole (

r = 0

). Because

r

is a distance, watch for angles where

f(\theta)

would be negative: under the

r \geq 0

convention those angles contribute no point.

Symmetry saves work. If

f(-\theta) = f(\theta)

the curve is symmetric about the initial line, so you need only sketch

0 \leq \theta \leq \pi

and reflect; if

f(\pi - \theta) = f(\theta)

it is symmetric about the line

\theta = \tfrac{\pi}{2}

. Spotting a symmetry first both halves the tabulation and guards against a lopsided sketch.

A handful of standard curves recur. The cardioid

r = a(1 + \cos\theta)

is a heart-shaped curve with a cusp at the pole and maximum radius

2a

along the initial line. The rose

r = a\cos(n\theta)

(or

a\sin n\theta

) has

n

petals when

n

is odd and

2n

petals when

n

is even, each petal reaching out to radius

a

. The spiral

r = a\theta

winds outward at a steady rate. Recognising these forms from the equation lets you sketch immediately.

The greatest distance from the pole occurs where

\dfrac{\mathrm{d}r}{\mathrm{d}\theta} = 0

with

r

a maximum, which locates the tips of petals and the far point of a cardioid. Marking these extreme points, the pole-crossings and any symmetry produces an accurate sketch. A good sketch is not merely decorative: the limits of the area integral in the final section are read directly from where the curve starts and finishes a loop, so an incorrect sketch usually leads to incorrect limits.

r = a(1 + \cos\theta) \ (\text{cardioid}), \qquad r = a\cos n\theta \ (\text{rose}), \qquad r = a\theta \ (\text{spiral})

Standard polar curves

A rose has $n$ petals for odd $n$ and $2n$ petals for even $n$ .

Worked example

Maximum radius of a limaçon

For the curve $r = 1 + 2\cos\theta$ , find the value of $\theta$ in $[0, \pi]$ giving the greatest $r$ , and the least value of $r$ that occurs.

Differentiate
$\dfrac{\mathrm{d}r}{\mathrm{d}\theta} = -2\sin\theta$ , which is zero at $\theta = 0$ and $\theta = \pi$ .
Evaluate at the stationary angles
$\theta = 0$ : $r = 1 + 2 = 3$ (maximum). $\theta = \pi$ : $r = 1 - 2 = -1$ .
Apply the r ≥ 0 convention
The curve reaches its greatest distance $r = 3$ along the initial line; where $1 + 2\cos\theta < 0$ (near $\theta = \pi$ ) there are no points, and the inner loop forms as $r$ returns to $0$ at $\cos\theta = -\tfrac12$ , i.e. $\theta = \tfrac{2\pi}{3}$ .
$r = 0 \ \text{at}\ \cos\theta = -\tfrac12,\ \theta = \tfrac{2\pi}{3}$
Conclude
Greatest $r = 3$ at $\theta = 0$ ; the curve touches the pole at $\theta = \tfrac{2\pi}{3}$ .

Result: Maximum $r = 3$ at $\theta = 0$ ; the curve reaches the pole at $\theta = \tfrac{2\pi}{3}$ .

Typical mistakes

Plotting points where $r$ would be negative, contrary to the $r\geq 0$ convention.
Miscounting the petals of a rose — $n$ petals for odd $n$ , $2n$ for even $n$ .

Active revision

Sketch the cardioid $r = 1 + \cos\theta$ for $0 \leq \theta < 2\pi$ , marking the maximum radius, the cusp and the symmetry.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Tangents to polar curves#

AdvancedAQA 7367 §3.7DfE Further Mathematics

To find tangents to a polar curve that are parallel or perpendicular to the initial line, convert to Cartesian coordinates through

x = r\cos\theta

and

y = r\sin\theta

, treating both as functions of

\theta

via

r = f(\theta)

. A tangent is parallel to the initial line where the curve is momentarily moving horizontally, i.e. where

\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = 0

; it is perpendicular to the initial line where the curve is moving vertically, i.e. where

\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = 0

Carrying out the differentiation,

y = f(\theta)\sin\theta

gives

\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = f'(\theta)\sin\theta + f(\theta)\cos\theta

, and similarly

\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = f'(\theta)\cos\theta - f(\theta)\sin\theta

. Setting the appropriate derivative to zero and solving for

\theta

locates the tangent points; substituting

\theta

back into

r = f(\theta)

gives the polar (or Cartesian) coordinates of each.

A subtlety arises at the pole. Where

r = 0

the tangent to the curve is the line

\theta = \alpha

for the angle

\alpha

at which the curve reaches the pole, provided

f'(\alpha) \neq 0

. This gives the direction in which a petal or loop leaves the origin, and is often asked for alongside the parallel/perpendicular tangents to complete the description of the curve's shape near the pole.

These tangent calculations pin down the exact shape of a curve: the horizontal tangents give the topmost and bottommost points of a loop, and the vertical tangents its left and right extremes. On a cardioid

r = a(1 + \cos\theta)

, for instance, solving

\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = 0

locates the highest point of the upper half. Combined with the sketch, they make the drawing quantitatively accurate rather than merely schematic.

\frac{\mathrm{d}y}{\mathrm{d}\theta} = f'(\theta)\sin\theta + f(\theta)\cos\theta = 0 \ (\text{tangent parallel to initial line})

Horizontal tangent

The curve is momentarily moving horizontally.

\frac{\mathrm{d}x}{\mathrm{d}\theta} = f'(\theta)\cos\theta - f(\theta)\sin\theta = 0 \ (\text{tangent perpendicular to initial line})

Vertical tangent

The curve is momentarily moving vertically.

Worked example

A horizontal tangent to a cardioid

Find the value of $\theta$ in $\left(0, \tfrac{\pi}{2}\right)$ where $r = 1 + \cos\theta$ has a tangent parallel to the initial line.

Form y in terms of θ
$y = r\sin\theta = (1 + \cos\theta)\sin\theta = \sin\theta + \sin\theta\cos\theta$ .
Differentiate
$\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = \cos\theta + \cos^2\theta - \sin^2\theta = \cos\theta + \cos 2\theta = 2\cos^2\theta + \cos\theta - 1$ .
Solve = 0
$2\cos^2\theta + \cos\theta - 1 = 0 \Rightarrow (2\cos\theta - 1)(\cos\theta + 1) = 0$ , so $\cos\theta = \tfrac12$ or $\cos\theta = -1$ .
$(2\cos\theta - 1)(\cos\theta + 1) = 0$
Select the root in range
In $\left(0, \tfrac{\pi}{2}\right)$ , $\cos\theta = \tfrac12$ gives $\theta = \tfrac{\pi}{3}$ (the root $\cos\theta = -1$ gives $\theta = \pi$ , the cusp).

Result: The tangent is parallel to the initial line at $\theta = \tfrac{\pi}{3}$ .

Typical mistakes

Confusing the conditions — using $\dfrac{\mathrm{d}x}{\mathrm{d}\theta}=0$ for a tangent parallel to the initial line instead of $\dfrac{\mathrm{d}y}{\mathrm{d}\theta}=0$ .
Forgetting to apply the product rule to $r\sin\theta$ and $r\cos\theta$ , since $r$ itself depends on $\theta$ .

Active revision

For the cardioid $r = 1 + \cos\theta$ , find the value of $\theta$ in $\left(0, \tfrac{\pi}{2}\right)$ at which the tangent is parallel to the initial line.

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Area enclosed by a polar curve#

AdvancedAQA 7367 §3.7DfE Further Mathematics

The four-petal rose r = 2 cos 2θ

Fig. 3One petal is traced between consecutive zeros of

r

; its area is

\tfrac{1}{2}\int (2\cos 2\theta)^2\,\mathrm{d}\theta = \tfrac{\pi}{2}

The area swept out by the radius as

\theta

increases is found not with strips but with thin sectors. A small change

\mathrm{d}\theta

sweeps a sector of radius

r

and angle

\mathrm{d}\theta

, whose area is approximately

\tfrac{1}{2}r^2\,\mathrm{d}\theta

(the area of a circular sector). Summing these gives the polar area formula

A = \dfrac{1}{2}\displaystyle\int_{\alpha}^{\beta} r^2\,\mathrm{d}\theta

, where

\alpha

and

\beta

are the angles that bound the region.

Choosing the limits correctly is where the sketch pays off:

\alpha

and

\beta

are the angles at which the boundary of the region begins and ends. For a full closed curve such as a cardioid, the limits are

0

and

2\pi

; for a single petal of a rose, they are the two consecutive angles at which

r

returns to zero, so the petal is traced exactly once. Integrating over the wrong range double-counts or misses part of the area.

Evaluating

\int r^2\,\mathrm{d}\theta

almost always requires the double-angle identities to handle

\cos^2

and

\sin^2

terms, since

\cos^2\theta = \tfrac{1}{2}(1 + \cos 2\theta)

and

\sin^2\theta = \tfrac{1}{2}(1 - \cos 2\theta)

. Expanding

r^2

first, then linearising every squared trigonometric term, turns the integral into a routine one. This is the same linearisation used with de Moivre's theorem, tying the topics together.

Areas between two polar curves, or between a curve and a line, are found by subtracting the appropriate sector-area integrals, exactly as areas between Cartesian curves are found by subtracting. The one extra care is to identify the angles at which the curves intersect, which set the limits, by solving

r_1(\theta) = r_2(\theta)

. A well-labelled sketch showing which curve is the outer boundary over each angular range is essential to getting the subtraction the right way round.

A = \frac{1}{2}\int_{\alpha}^{\beta} r^2\,\mathrm{d}\theta

Area in polar coordinates

Sum of thin sectors of area $\tfrac{1}{2}r^2\,\mathrm{d}\theta$ between the bounding angles.

Worked example

Area enclosed by a cardioid

Show that the area enclosed by $r = 1 + \cos\theta$ is $\tfrac{3\pi}{2}$ .

Set up the integral
$A = \dfrac{1}{2}\displaystyle\int_0^{2\pi} (1 + \cos\theta)^2\,\mathrm{d}\theta$ .
Expand
$(1+\cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta = 1 + 2\cos\theta + \tfrac{1}{2}(1 + \cos 2\theta) = \tfrac{3}{2} + 2\cos\theta + \tfrac{1}{2}\cos 2\theta$ .
Integrate
$A = \dfrac{1}{2}\left[\tfrac{3}{2}\theta + 2\sin\theta + \tfrac{1}{4}\sin 2\theta\right]_0^{2\pi}$ .
$A = \frac{1}{2}\left[\tfrac{3}{2}\theta + 2\sin\theta + \tfrac{1}{4}\sin 2\theta\right]_0^{2\pi}$
Evaluate
The sine terms vanish at $0$ and $2\pi$ , leaving $A = \dfrac{1}{2}\cdot\tfrac{3}{2}(2\pi) = \dfrac{3\pi}{2}$ .

Result: $A = \dfrac{3\pi}{2}$ .

Typical mistakes

Forgetting the factor $\tfrac{1}{2}$ , or integrating $r$ rather than $r^2$ .
Using the wrong limits, so a petal is traced more than once or a region is only partly covered.

Active revision

Show that the total area enclosed by the cardioid $r = 1 + \cos\theta$ is $\tfrac{3\pi}{2}$ , and find the area of one petal of the rose $r = 2\cos 2\theta$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

References & sources

Sources

Department for Education

Further mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Further Mathematics 7367 specification

Polar coordinates

Key points

Converting a polar equation

Key points

Maximum radius of a limaçon

Key points

A horizontal tangent to a cardioid

Key points

Area enclosed by a cardioid

Sources